我花了一天时间试图理解python类模型的复杂性，弄乱了装饰器，元类和超类。
目前，我正在试图找出某些令牌功能的作用，即新（这里的背景故事Metaclasses and when/how functions are called）
我已经制作了一个新的模拟模块来运行测试，在这里：
#! /usr/bin/env python3
import sys as system
import os as operating_system
from functools import partial
from time import perf_counter as counter
class Meta(type):
@classmethod
def __prepare__(instance, name, supers, *list, **map):
print('{} in meta prepare'.format(name))
return {}
def __new__(instance, name, supers, attributes, *list, **map):
print('{} in meta new'.format(name))
return instance
def __init__(self, name, supers, attributes, *list, **map):
print('{} in meta init'.format(self))
def __call__(self, *list, **map):
print('{} in meta call'.format(self))
return type.__call__(self)
print('after call')
class Super(object):
def __new__(instance, *list, **map):
print('{} in Super new'.format(instance))
return instance
def __init__(self, *list, **map):
print('{} in Super init'.format(self))
def __call__(self, *list, **map):
print('{} in Super call'.format(self))
return object.__call__(self)
class Other(object):
def __new__(instance, *list, **map):
print('{} in Other new'.format(instance))
return instance
def __init__(self, *list, **map):
print('{} in Other init'.format(self))
def __call__(self, *list, **map):
print('{} in Other call'.format(self))
return object.__call__(self)
class MetaSuper(object, metaclass = Meta):
def __new__(instance, *list, **map):
print('{} in MetaSuper new'.format(instance))
return instance
def __init__(self, *list, **map):
print('{} in MetaSuper init'.format(self))
def __call__(self, *list, **map):
print('{} in MetaSuper call'.format(self))
return object.__call__(self)
class DoubleSuper(Super, MetaSuper):
def __new__(instance, *list, **map):
print('{} in DoubleSuper new'.format(instance))
return instance
def __init__(self, *list, **map):
print('{} in DoubleSuper init'.format(self))
Super.__init__(self, *list, **map)
MetaSuper.__init__(self, *list, **map)
def __call__(self, *list, **map):
print('{} in DoubleSuper call'.format(self))
return object.__call__(self)
class SuperThenMeta(Super, metaclass = Meta):
def __new__(instance, *list, **map):
print('{} in SuperThenMeta new'.format(instance))
return instance
def __init__(self, *list, **map):
print('{} in SuperThenMeta init'.format(self))
Super.__init__(self, *list, **map)
def __call__(self, *list, **map):
print('{} in SuperThenMeta call'.format(self))
return object.__call__(self)
class Triple(Super, Other, metaclass = Meta):
def __new__(instance, *list, **map):
print('{} in Triple new'.format(instance))
return instance
def __init__(self, *list, **map):
print('{} in Triple init'.format(self))
Super.__init__(self, *list, **map)
Other.__init__(self, *list, **map)
def __call__(self, *list, **map):
print('{} in Triple call'.format(self))
return object.__call__(self)
class Simple(Super):
def __new__(instance, *list, **map):
print('{} in Simple new'.format(instance))
return instance.__init__(instance, *list, **map)
def __init__(self, *list, **map):
print('{} in Simple init'.format(self))
Super.__init__(self, *list, **map)
Other.__init__(self, *list, **map)
def __call__(self, *list, **map):
print('{} in Simple call'.format(self))
return object.__call__(self)
def main():
#thing = SuperThenMeta()
#other = DoubleSuper()
last = Super()
simp = Simple()
trip = Triple()
if __name__ == '__main__':
main()
TL; DR，我在这些工作片之间尝试了几种不同的设置。
如果我运行它，这是输出：
MetaSuper in meta prepare
MetaSuper in meta new
SuperThenMeta in meta prepare
SuperThenMeta in meta new
Triple in meta prepare
Triple in meta new
<class '__main__.Super'> in Super new
<class '__main__.Simple'> in Simple new
<class '__main__.Simple'> in Simple init
<class '__main__.Simple'> in Super init
<class '__main__.Simple'> in Other init
Traceback (most recent call last):
File "./metaprogramming.py", line 134, in <module>
main()
File "./metaprogramming.py", line 131, in main
trip = Triple()
TypeError: __new__() missing 3 required positional arguments: 'name', 'supers', and 'attributes'
由此，我有几个问题：
我应该在新函数末尾调用实例。 init （实例，*列表，**地图）？我不这么认为，但在“简单”示例中添加它似乎有效，而“超级”从未到达 init 。我的印象是通过在我自己的调用方法中调用object。调用，这将由它的默认实现处理，但在整个程序期间没有__call__s。
为什么调用Triple（）首先调用元类 new ？如果这是正常的，这是否意味着这是具有元类的任何类的典型？这是与超类类似的行为吗？
我希望调用在此列表中。是否在对象的创建例程中调用它（例如[prepare]，new，init）？
我知道这是很多信息，所以感谢你阅读这篇文章;任何指导都将不胜感激。
答案 0 :(得分：1)
__new__ 方法__new__是创建新实例的方法。因此它的第一个参数不是和实例，因为还没有创建它，而是类本身。
对于元类，__new__应该返回元类的实例，即类。它的签名是这样的：
class Meta(type):
def __new__(metacls, name, bases, namespace, **kwargs):
...
metacls是元类本身。
name是一个表示所在类名称的字符串
实例
bases是类将继承的类的元组。
namespace是类的名称空间，这是对象
由__prepare__返回，现在填充了类属性。
**kwargs是在实例化时传递给类的任何关键字参数
要实例化一个类，您需要调用type.__new__，这是默认的元类。您通常可以致电super().__new__。
class Meta(type):
def __new__(metacls, name, bases, namespace, **kwargs):
print('You can do stuff here')
cls = super().__new__(metacls, name, bases, namespace, **kwargs)
# You must return the generated class
return cls
__init__ __init__方法与任何其他类的行为没有区别。如果__new__返回了期望类型的实例，它将接收创建的实例（此处为类）作为参数。在您的示例中，__new__不会返回Meta类型的对象。它返回Meta本身，其类型为type。
在实例化中永远不会调用以下__init__方法。
class Meta(type):
def __new__(metacls, name, bases, namespace, **kwargs):
return None # or anything such that type(obj) is not Meta
def __init__(self, name, bases, namespace, **kwargs):
# This will never be called because the return type of `__new__` is wrong
pass
在实例化时调用以下内容，因为Meta.__new__正确返回类型为Meta的对象。
class Meta(type):
def __new__(metacls, name, bases, namespace, **kwargs):
return super().__new__(metacls, name, bases, namespace, **kwargs)
def __init__(self, name, bases, namespace, **kwargs):
print('__init__ was called')
__call__ 同样，__call__的行为与任何其他类别的行为没有什么不同。当您尝试调用元类的实例时调用它，而在调用元类创建时调用__new__和__init__实例（一个类）。
当然，调用一个类应该返回一个实例，所以不要忘记调用super().__call__并返回其结果，否则你将短路实例创建。
class Meta(type):
def __call__(self, *args, **kwargs):
print(f'An instance was called with {args}')
return super().__call__(self, *args, **kwargs)
# This declaration if what calls __new__ and __init__ of the metaclass
class Klass(metaclass=Meta):
pass
# This calls the __call__ method of the metaclass
instance = Klass()